[leetcode] 74. Search a 2D Matrix 解题报告
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题目链接:https://leetcode.com/problems/search-a-2d-matrix/
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
- Integers in each row are sorted from left to right.
- The first integer of each row is greater than the last integer of the previous row.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
Given target = 3
, return true
.
思路:昨天是跨年,白天给davis教授搬了一天家,他们一家就要搬到弗吉尼亚了,很nice的老师。晚上和同学一起到greenville的中国餐馆吃了顿饭,然后回到UV一起玩了会牌在午夜12的一起放了烟花拍了照片,完了就差不多晚上两点了,感觉好累,就睡觉了,所以一天都没有刷题。好吧,吐槽了几句现在的生活,anyway,happy new year!
这题的思路很简单,利用矩阵行列分别有序的性质,我们从第0行的最右端开始搜索,两种情况:
1. 如果当前的数比target大,则说明target必然在target的左边,则搜索其左边的位置
2. 如果当前的数比target小,则说明和target相等的数必然不在这一行,因此往下一行搜索。
3. 如果当前的数和target相等,则说明找到了这个target
代码如下:
class Solution {public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int i = 0, j = matrix[0].size() -1; while(i < matrix.size() && j >= 0) { if(matrix[i][j] == target) return true; else if(matrix[i][j] > target) j--; else if(matrix[i][j] < target) i++; } return false; }};
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